• Scale physical model • Mathematical model • Numerical model How to Model Physical Systems www.autospeed.com
Dec 23, 2015
• Scale physical model
• Mathematical model
• Numerical model
How to Model Physical Systems
www.autospeed.com
• Develop mathematical models, i.e. ordinary differential equations that describe the relationship between input and output characteristics of a system.
• These equations can then be used to forecast the behaviour of the system under specific conditions.
• All systems can normally be approximated and modelled by one of several models, e.g. mechanical, electrical, thermal or fluid. We also find that we can translate a system from one model to another to facilitate the modelling.
Modelling of Physical Systems
Lumped Parameter Models
• Use standard laws of physics and break a system down into a number of building blocks.
• Each of the parameters (property or function) is considered independently.
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Lumped Parameter Models
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Linear Time Invariant Models
• Assume the property of linearity for these models.
• A linear system will posses two properties;
1. Superposition
2. Homogeneity.
Allows us to use standard mathematical operations to simplify our models
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Linear Time Invariant Models
• Assume system is time-invariant
• Constants stay constant in the time-scales of our model
• Proportionality between variables does not change.
Our shock absorbers do not wear in our car suspension model!
Competition tyres
Elements of Systems are Ideal • Each element completely describes a property
• Elements are:
• Ideal
• Linear
• Represent only one property
The Spring ElementRepresents the elastic properties (energy storage) in a system.
Assume:
• No mass
• No Fraction
• Linear
Hooke’s Law: f(t) = K(x1 – x2) = Kx(t)
f(t) = force applied to the ends of the spring, K = spring constant (N.m) or C = 1/K is the spring compliance,x1 = Displacement of the one end, x2 = Displacement of the other end,
x = x1 – x2 = relative displacement of the two ends. Rotational spring element torque T(t) in terms of the angular displacement :
T(t) = K(1 - 2) = K(t) T(t) = Torque(t) = angular displacement
The Spring Element
Real springs show deviation from ideal:
• Real springs would have an associated mass, leading to the deviations in the step and sinusoidal response as described before.
• Real springs will always contain some friction (energy dissipation) which is exhibited in the non-coincidence of the loading – unloading curves.
• Real springs will always exhibit a degree of non-linearity (deviation from f = Kx). See section on linearisation.
The Spring Element
Linearisation Many elements show non-linear behaviour.
To use in our relatively simple models, we must first linearise this element around an operating point.
Get a linear model that should be valid for small excursions from this operating point.
Consider a non-linear coil spring that forms part of the suspension system of an automobile.
Operating point = Equilibrium position under load
Example: Consider f(x) = 2x + 5x3
f(x) = f(2 (xo)+5(xo)3) + f’(2 +3.5 xo 2).(x-1)
f(x) = f(2 (1)+5(1)3) + (2 +3.5.12)(x-1)
Linearise: Intuitively with tangent
Or Taylor series:
A finite number of terms will give an approximation of the function – the first two terms will give a linear approximation
Example: Consider f(x) = 2x + 5x3
Operating point x = 1: f(x) ~ -10 + 17x
Model the spring force as f(x) = -10 + 17x around the point x = 1.0 and
the linearised spring forced constant would be given by df/dx = 17 N/m.
Linearisation
...
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2
2
00
00
xx
dx
ydxx
dx
dyxfy
xx
The Damper ElementDamper element (dashpot) represents the forces opposing motion, i.e. the friction or damping effects.
Damping force is proportional to the velocity of the piston.
where f = force applied to the ends of the damper,
dx/dt = v = relative velocity of the plunger
fv = coefficient of viscous friction or simply called the damping coefficient
vfdt
dxf
dt
dx
dt
dxff vvv
21
The Damper Element
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The Mass (inertia) ElementAll real elements will have some mass, so that the mass element in the model will thus represent the physical mass of the system.
This element represents the inertia or resistance to acceleration of the system.
The mass element will be modelled as concentrated at a point.
The three solutions • stiffening the structure, • dampening structural vibration, • limiting/modifying bridge traffic Estimates place the cost of the stabilization work in the neighborhood of £5 million.(Source: http://pub20.ezboard.com/fglobalarchitectureforumsother.showMessage?topicID=145.topic)
Millennium Bridge London
Car suspension model:
• Mechanical system
Explain what happens when a car goes over a bump?
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Car suspension model:Explain what happens when a car goes over a bump?
x
finput
Simplify to single-input-single-output system
Form individual component models
Determine their relationships (use physical laws!)
Combine (and simplify if possible)
This gives us an instantaneous differential equation, but want a time response!
Spring
Damper
Mass
Source Nise 2004
Force - Distance
tKxtf s
dt
tdxCtfc
2
2
dt
txdMtfm
tf
tx
tf
tx
tf
tx
tftftftf mdsinput
Car suspension model:
2
2
dt
txdm
dt
tdxCtKxtfinput
This gives us an instantaneous differential equation, but want a time response!
Integrate:• numerically• theoretically• using tables
clf; %clear all graphs K = 10 %Spring constantC = 3 %Damping constantm = 1 %mass (constant)
t = [0: 0.01: 20];%set up the time incrementsstept = 1 + 0*t; %graph to show step responseplot(t,stept,'m');xlabel('Time t (s)')ylabel('Distance x (m)')
hold on % put each graph on top of each other
for C = 1.0: 1: 10.0d = tf(9,[m C K]) [y,t]=step(d,T);%step response over one secondplot(t,y,'k');pause(2)
end
0 2 4 6 8 10 12 14 16 18 200
1
2
3
4
5
6
7
8
9
Time t (s)
Dis
tanc
e x
(m)
0 2 4 6 8 10 12 14 16 18 200
0.5
1
1.5
Time t (s)
Dis
tanc
e x
(m)
0 2 4 6 8 10 12 14 16 18 200
0.5
1
1.5
Time t (s)
Dis
tanc
e x
(m)